Question: $\dfrac{ 6e - 2f }{ -10 } = \dfrac{ 3e + 3g }{ 2 }$ Solve for $e$.
Solution: Multiply both sides by the left denominator. $\dfrac{ 6e - 2f }{ -{10} } = \dfrac{ 3e + 3g }{ 2 }$ $-{10} \cdot \dfrac{ 6e - 2f }{ -{10} } = -{10} \cdot \dfrac{ 3e + 3g }{ 2 }$ $6e - 2f = -{10} \cdot \dfrac { 3e + 3g }{ 2 }$ Reduce the right side. $6e - 2f = -{10} \cdot \dfrac{ 3e + 3g }{ {2} }$ $6e - 2f = -{5} \cdot \left( 3e + 3g \right)$ Distribute the right side $6e - 2f = -{5} \cdot \left( {3e} + {3g} \right)$ $6e - 2f = -{15}e - {15}g$ Combine $e$ terms on the left. ${6e} - 2f = -{15e} - 15g$ ${21e} - 2f = -15g$ Move the $f$ term to the right. $21e - {2f} = -15g$ $21e = -15g + {2f}$ Isolate $e$ by dividing both sides by its coefficient. ${21}e = -15g + 2f$ $e = \dfrac{ -15g + 2f }{ {21} }$